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The laws of Physics Topics are used to explain everything from the smallest subatomic particles to the largest galaxies.

## Relations Among the Elastic Constants

Relations among Young’s modulus (Y), bulk modulus (K), modulus of rigidity (n) and Poisson’s ratio (σ) are shown below.

These relations show that any two of the four quantities are independent. If the values of any two quantities are known, then the other two can be found out.

### Work Done In Stretching A Wire : Energy Density

When a body gets strained, an internal reaction force develops inside the body. As a result, the applied external force does some work against this reaction force to produce deformation in the body. This work remains stored inside the body as potential energy, known as elastic potential energy. If the external force is withdrawn, then the internal reaction force, i.e., stress, ceases and this potential energy is converted into heat energy.

Let us consider a wire of length L and cross sectional area α, whose one end is fixed to a rigid support aid a force F is applied at its other end. Let the total elongation of the wire be l. Within the elastic limit, the elongation is directly proportional to the applied force. When the applied force is zero, the elongation is zero; and when the applied force is F,

then the elongation of the wire becomes l. So we can say that during the elongation l of the wire, an average force \(\frac{0+F}{2}\) or \(\frac{F}{2}\) actually acts on the wire.

Therefore, work done,

W = average force × displacement = \(\frac{1}{2}\)Fl ……. (1)

The Young’s modulus for the material of the wire is,

Y = \(\frac{F / \alpha}{l / L}\), or F = \(\frac{Y \alpha l}{L}\) ∴ W = \(\frac{1}{2}\)Fl = \(\frac{1}{2} \frac{Y \alpha l^2}{L}\) …. (2)

This work remains stored in the wire as potential energy

Volume of the wire = Lα.

Therefore, the work done per unit volume of the wire, or the potential energy stored per unit volume of the wire (energy density), is

So, elastic energy density = \(\frac{1}{2}\) × stress × strain

Calculation with the he of calculus: If the total elongation of the wire is supposed to be an aggregate of infinitesimal elongations, then for an elongation dl of the wire, the work done, dW = Fdl

We know that, F = \(\frac{Y \alpha l}{L}\) ∴ dW = \(\frac{Y \alpha l}{L} \cdot d l\)

Therefore, the work done for the total elongation l becomes,

This is exactly same as equation (2).

### Numerical Examples

**Example 1.**

What amount of work must be done in stretching a wire, of length 1 m and of cross sectional area 1 mm^{2}, by 0.1 mm? Youngs modulus for the material = 2 × 10^{11} N ᐧ m^{-2}.

**Solution:**

We know that, work done, W = \(\frac{1}{2} \frac{Y \alpha l^2}{L}\)

Here, Y = 2 × 10^{11} N ᐧ m^{-2}, α = 1 mm^{2} = 10^{-6} m^{2},

l = 0.1 mm = 10^{-4}m,L=lm ‘

∴ W = \(\frac{1}{2} \times \frac{2 \times 10^{11} \times 10^{-6} \times\left(10^{-4}\right)^2}{1}\) = 10^{-3}J.

**Example 2.**

When the load on a wire is increased from 3 kg to 5 kg, the elongation of the wire increases from 0.6 mm to 1 min. How much work is done during this extension of the wire?

**Solution:**

We know that, work done, W = \(\frac{1}{2}\)Fl.

Let the work done in stretching the wire through 0.6 mm be W_{1}.

Here, F = 3 kgf = (3 × 9.8) N

and l = 0.6 mm = 6 × 10^{-4}m

∴ W_{1} = \(\frac{1}{2}\) × 3 × 9.8 × 6 × 10^{-4} = 8.82 × 10^{-3}J

Let the work done in stretching the wire through 1 mm be W_{2}.

In this case, F = 5 kgf = 5 × 9.8 N

and I = 1 mm = 10^{-3} m

∴ W_{2} = \(\frac{1}{2}\) × 5 × 9.8 × 10^{-3} = 2.45 × 10^{-2} J

So, during the extension of the wire from 0.6 mm to 1 mm, the net work done = W_{2} – W_{1} = (2.45 – 0.882) × 10^{-2}

= 1.568 × 10^{-2} J

**Example 3.**

For a uniform wire of length 3 m and cross sectional area 1 mm^{2}, 0.021 J of work is necessary to stretch it through 1 mm. Calculate the Young’s modulus for its material.

**Solution:**

Work done, W = \(\frac{1}{2} \frac{Y \alpha l^2}{L}\)

or, Y = \(\frac{2 W L}{\alpha l^2}\)

Here, W = 0.021 J, α = 1 mm^{2} = 10^{-6}m^{2}, L = 3 m, l = 1 mm = 10^{-3} m

∴ Y = \(\frac{2 \times 0.021 \times 3}{10^{-6} \times\left(10^{-3}\right)^2}\) = 1.26 × 10^{11} N ᐧ m^{-2}

**Example 4.**

Two uniform wires of length 3 m and 4 m respectively are made of the same material. To stretch both these wires by the same length, 0.03 J and 0.05 J of work are necessary. Calculate the ratio of the cross sectional areas of the two wires.

**Solution:**

Let the work done in the case of the first and the second wires respectively be,

W_{1} = \(\frac{1}{2} \frac{Y \alpha_1 l^2}{L_1}\) and W_{2} = \(\frac{1}{2} \frac{Y \alpha_2 l^2}{L_2}\)

∴ \(\frac{W_1}{W_2}\) = \(\frac{\alpha_1 L_2}{\alpha_2 L_1}\) or, \(\frac{\alpha_1}{\alpha_2}\) = \(\frac{W_1 L_1}{W_2 L_2}\)

Here, W_{1} = 0.03J, W_{2} = 0.05J, L_{1} = 3 m and L_{2} = 4 m

∴ \(\frac{\alpha_1}{\alpha_2}\) = \(\frac{0.03 \times 3}{0.05 \times 4}\) or, α_{1} : α_{2} = 9 : 20